Question

Two charges +20micro coulomb and 90micro coulomb placed apart 18cm .find the point where electric field is zero?

Answer 1

Answer:

Given:

+20 micro-C and +90 micro-C charge has been placed at a separation of 18 cm.

To find:

Position where the Electric Field is zero.

Concept:

There will be a position on the line joining the 2 charges where the electric Field Intensity of both the charges will be equal and opposite to one another and hence will cancel our each other.

That point is also known as the Neutral Point .

Calculation:

Let the point be x distance from 20 micro-C.

At the neutral point , we can say that the electric Field Intensity by both charges will be equal :

$E1 = E2$

$= > \dfrac{k(20 \times {10}^{ - 6}) }{ {x}^{2} } = \dfrac{k(90 \times {10}^{ - 6}) }{ {(18 - x)}^{2} }$

$= > \dfrac{2}{ {x}^{2} } = \dfrac{9}{ {(18 - x)}^{2} }$

$= > \dfrac{ \sqrt{2} }{x} = \dfrac{3}{(18 - x)}$

$= > 18 \sqrt{2} - x\sqrt{2} = 3x$

$= > x(3 + \sqrt{2} ) = 18 \sqrt{2}$

$= > x = \dfrac{18 \sqrt{2} }{(3 + \sqrt{2}) } cm$

So position of the point will be as calculated above from 20 micro-C

So final Answer :

$\boxed{ \red{ \huge{ \bold{ x = \dfrac{18 \sqrt{2} }{(3 + \sqrt{2}) } cm}}}}$

Answer 2

$\mathtt{ \huge{\fbox{Solution :)}}}$

Given ,

• The two charges are 20 micro C or 20 × (10)^-6 C and 90 micro C or 90 × (10)^-6 C

• The distance between two charges is 18 cm

Let ,

• P be the point where electric field is zero

• The distance of first charge from point P be x cm

• Then , the distance of second charge from point P is (18 - x) cm

The electric field at point P will be zero if

Electric field due to first charge = Electric field due to second charge

We know that , the ratio of electronstatic force per unit test charge is called electric field

$\mathtt{ \large{ \fbox{Electric \: field = \frac{1}{4\pi e_{0}} \frac{q}{ {(r)}^{2} } }}}$

Thus ,

$\sf \mapsto \frac{1}{4 \pi e_{o }} \frac{20 \times {(10)}^{ - 6} }{ {(x)}^{2} } = \frac{1}{4 \pi e_{o }} \frac{90 \times {(10)}^{ - 6} }{ {(18 - x)}^{2} } \\ \\ \sf \mapsto {(18 - x)}^{2} = \frac{9}{2} {(x)}^{2} \\ \\ \sf \: \: Squaring \: both \: sides \: , \: we \: get \\ \\ \sf \mapsto18 - x = \frac{3x}{ \sqrt{2} } \\ \\ \sf \mapsto 18\sqrt{2} - \sqrt{2}x = 3x \\ \\ \sf \mapsto 18 \sqrt{2} = 3x + \sqrt{2}x \\ \\\sf \mapsto 18 \sqrt{2} = x(3 + \sqrt{2} ) \\ \\\sf \mapsto x = \frac{18 \sqrt{2} }{3 + \sqrt{2} } \: \: cm$

Hence , electric field will be zero at $\mathtt{ \large{ \fbox{</u></strong><strong><u> </u></strong><strong><u> \frac{18 \sqrt{2} }{3 + \sqrt{2} } </u></strong><strong><u>}}}$ cm to the right of first charge

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